Measure and Integration: Lecture 9 Notes

MEASURE AND INTEGRATION: LECTURE 9 Invariance of Lebesgue measure. Given A ⊂ Rn and z ∈ Rn , let z + A = {z + x | x ∈ A} be the translate of A by z. Given t > 0, let tA = {tx | x ∈ A} be the dilation of A by t. Let I = [a1 , b1 ] × · · · × [an , bn ] and z = z1 × · · · × zn . Then z + I = [z1 + a1 , z1 + b1 ] × · · · × [zn + an , zn + bn ], and tI = [ta1 , tb1 ] × · · · × [tan , tbn ], and we have λ(z + I) = (z1 + b1 − z1 − a1 ) · · · (zn + bn − zn − az ) = (b1 − a1 ) · · · (bn − an ) = λ(I). and λ(tI) = tn · λ(I). If P is a special polygon, then λ(z + P ) = λ(P ) and λ(tP ) = tn P . � Indeed, write P = N i=1 Ii and the proof is straightforward. If G is an open set, then λ(z + G) = λ(G) and λ(tG) = tn λ(G). We have λ(G) = sup{λ(P ) | P ⊂ G special polygon}, so λ(z + G) = sup{λ(P ) | P ⊂ z + G, P special polygon}. But P ⊂ G special polygon ⇐⇒ z + P ⊂ z + G special polygon. Since Lebesgue invariance holds for special polygons, it holds for open sets. Finally, by similar reasoning, it can be shown that a set A ⊂ Rn is measurable if and only if z + A is measurable if and only if tA is measurable, and λ(A) = λ(z + A), λ(tA) = tn λ(A). A non­measurable set E ⊂ Rn . Let Q be the set of rational num­ bers. For x ∈ R, consider x+Q = {x+q | q ∈ Q}. Then y ∈ x+Q ⇐⇒ y − x ∈ Q. Claim: if x, x� ∈ R, then either (i) x + Q = x� + Q or (ii) (x + Q) ∩ � (x + Q) = ∅. Proof: If the intersection is nonempty, then there exists y = x + q1 = x� + q2 , which implies that x − x� = q1 − q2 ∈ Q. Thus, x + Q = x� + Q, and the claim is proved. 1 2 MEASURE AND INTEGRATION: LECTURE 9 We have shown that R is covered disjointly by the sets x + Q. The Axiom of Choice states that there exists a set E ⊂ R such that every point of R belongs to only one of these sets, i.e., R= (x + Q) x∈E is a disjoint union. Alternatively, for any x ∈ R, there exists a unique y ∈ E and unique z ∈ Q such that x = y + z. Since the set Q is countable, its elements can be enumerated: Q = {q1 , q2 , . . .}. Thus, ∞ R= (qk + E) k=1 is a disjoint union. Using outer measure subadditivity and invariance of Lebesgue measure, ∗ λ (R) ≤ ∞ � ∗ λ (qk + E) = k=1 ∞ � λ∗ (E). k=1 ∗ Hence we must have that λ (E) > 0 (otherwise λ∗ (R) = 0). Now let K ⊂ E be an arbitrary compact subset of E and let D = (0, 1) ∩ Q. (The set D is a bounded countably infinite set.) Then (q + K) = D + K q∈D is a bounded set. The sets in the union are disjoint, since rational translates of E are disjoint. We have ∞ > λ(D + K) (bounded) � � =λ (q + K) q∈D = � λ(q + K) q∈D = � λ(K). q∈D Since the sum is over an infinite index set, λ(K) = 0. Because K ⊂ E arbitrary ⇒ λ(K) = 0, we have λ∗ (E) = 0. But 0 = λ( E) < λ∗ (E) ⇒ E �∈ L. Corollary 0.1. If A ⊂ Rn is measurable with positive measure, then there exists B ⊂ A that is not measurable. MEASURE AND INTEGRATION: LECTURE 9 3 Proof. Write A = ∪∞ k=1 ((qk + E) ∩ A) as a disjoint union. Then 0 < λ(A) = λ∗ (A) ≤ ∞ � λ∗ ((qk + E) ∩ A) , k=1 ∗ and so λ ((qk + E) ∩ A) > 0 for some k. But λ∗ ((qk + E) ∩ A) ≤ λ∗ (qk + E) = λ∗ (E) = 0, a contradiction. � Invariance under linear transformations. Theorem 0.2. Let T : Rn → Rn be a linear map and A ⊂ Rn . Then λ∗ (T A) = |det T | λ∗ (A), λ∗ (T A) = |det T | λ∗ (A). If A is measurable, then T A is measurable and λ(T A) = |det T | λ(A). Proof. First assume that T is invertible, i.e., that det T �= 0. We will use the following lemma. Lemma 0.3. Let T be invertible and let J = [0, 1)n . Let ρ be defined by λ(T J) = ρλ(J). If A ⊂ Rn , then λ∗ (T A) = ρλ∗ (A) and λ∗ (T A) = ρλ∗ (A). If A is measurable, then T A is measurable and λ(T A) = ρλ(A). Proof. The set J is the union of countably many compact sets: ∞ J= [0, 1 − 1/k]n , k=1 and so TJ = ∞ T ([0, 1 − 1/k]). k=1 Since T maps compact sets to compact sets, T J is the union of count­ ably many compact sets. Thus, T J is measurable, so the definition of ρ makes sense. We just to need to prove that λ(T G) = ρλ(G) for G open. As before, if the measure of open sets is invariant, then outer measure, compacts, and inner measures are invariant. Let G ⊂ Rn be open. Claim: can write G = ∪∞ k=1 Jk with Jk ’s disjoint and each Jk is a translation and dilation of J. (Pair by integer of those not contained, then pair by 1/2, then by 1/4, . . ..) Let Jk = zk + tk · J. Then λ(Jk ) = tnk λ(J). T Jk = T zk + tk · T J 4 MEASURE AND INTEGRATION: LECTURE 9 ⇒ λ(T Jk ) = tnk λ(T J) = tnk ρλ(J) = tnk ρt1−n λ(Jk ). k ∞ Thus, λ(T Jk ) = ρλ(Jk ). Since G = ∪k=1 Jk , T G = ∪∞ k=1 T Jk , which is a disjoint collection of measurable sets. Thus we have ∞ ∞ � � λ(T G) = λ(T Jk ) = ρ · λ(Jk ) = ρ · λ(G). k=1 k=1 � To identify ρ, check for elementary matrices just on the cube. This shows that in fact ρ = |det T |. Lastly, if T is not invertible, i.e., det T = 0, then the image T Rn is the subset of a hyperplane. This means that T A has measure zero, so the formula still holds. � A linear transformation is a rotation when the matrix is an orthogo­ nal matrix: AAT = I. In this case, it must be that det A = ±1. Thus, Lebesgue measure is invariant under rotation. Finally, there is an important subgroup of the group of all n × n real matrices known as the special linear group, denoted SL(n, R) = {A | det A = 1}.